# Product rule for differentiation

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Statement for two functions

### Verbal statement

If two (possibly equal) functions are differentiable at a given real number, then their pointwise product is also differentiable at that number and the derivative of the product is the sum of two terms: the derivative of the first function times the second function and the first function times the derivative of the second function.

### Statement with symbols

Suppose $f$ and $g$ are functions, both of which are differentiable at a real number $x = x_0$. Then, the product function $f \cdot g$, defined as $x \mapsto f(x)g(x)$ is also differentiable at $x$, and the derivative at $x_0$ is given as follows:

$\! \frac{d}{dx} [f(x)g(x)]|_{x = x_0} = f'(x_0)g(x_0) + f(x_0)g'(x_0)$

or equivalently:

$\! \frac{d}{dx} [f(x)g(x)]|_{x = x_0} = \frac{d(f(x))}{dx}|_{x=x_0} \cdot g(x_0) + f(x_0)\cdot \frac{d(g(x))}{dx}|_{x = x_0}$

If we consider the general expressions rather than evaluation at a particular point $x_0$, we can rewrite the above as:

$\! \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$

or equivalently:

$(f \cdot g)' = (f' \cdot g) + (f \cdot g')$